showing that f is continuous. Thus the sequence of functions converges uniformly to a function f ∈ C0

(X)

which is what it means to be complete. Every Cauchy sequence converges. Indeed 2.19 says that

∥f − f ∥
n

∞< ε for f ∈ C0

(X)

. ■

The above refers to functions which have values in ℂ but the same proof works for functions which have
values in any complete normed linear space.

In the case where the functions in C0

(X )

all have real values, I will denote the resulting space by
C0

(X;ℝ )

with similar meanings in other cases.

The following has to do with a trick which will enable a result valid on sets which are only closed rather
than compact. In general, you consider a locally compact Hausdorff space instead of a closed subset of ℝn.
Consider the unit sphere in ℝn+1, centered at the point

(0,⋅⋅⋅,0,1)

≡

(⃗ )
0,1

.

{ ∑n }
Sn ≡ ⃗x ∈ ℝn+1 : (xn+1 − 1)2 + x2k = 1
k=1

Define a map from ℝn which is identified with ℝn×

{0}

to the surface of this sphere as follows. Extend a
line from the point,

⃗p

in ℝn to the point

(⃗0,2)

on the top of this sphere and let θ

(p)

denote the point of
this sphere which the line intersects.

PICT

This map θ is one to one onto Sn∖

{( )}
⃗0,2

. More precisely, if you have

(⃗a,an+1)

on Sn∖

( )
⃗0,2

to
get θ−1

(⃗a,an+1)

, you form the line from

( )
⃗0,2

through this point and see where it hits ℝn. The line is

( )
⃗0,2

+ t

( ( ) )
(⃗a,a )− ⃗0,2
n+1

and it hits ℝn when 2 + t

(a − 2)
n+1

= 0 which is when t =

---2--
2−an+1

. Thus
θ−1

(⃗a,an+1)

=

( )
2−2a⃗an+1,0

. From this formula, it is clear that θ−1 is continuous and one to one. It is also
onto because if

⃗x

∈ ℝn you can take the line from

(⃗ )
0,2

to

(⃗x,0)

and where it intersects Sn is the point
which is wanted. It is also easy to see from this that θ is continuous. Indeed, suppose

⃗x

k→

⃗x

in ℝn. Does it
follow that θ

(⃗xk)

→ θ

(⃗x)

? We know that

{⃗xk }

is bounded since it converges. Therefore, there is an open
ball, B

( ( ) )
⃗0,2 ,r

such that θ

(⃗xk)

∈ Sn∖ B

(( ) )
⃗0,2 ,r

≡ K a compact set. If θ

(⃗xk)

fails to
converge to θ

(⃗x)

, then there is a subsequence, still denoted as θ

(⃗xk)

such that θ

(⃗xk)

→ y ∈ K
where y≠θ

(⃗x)

. But then, the continuity of θ−1 implies xk→ θ−1

(y)

and so θ−1

(y)

= x which
implies y = θ

(x )

, a contradiction. Thus both θ and θ−1 are continuous, one to one and onto
mappings.

Note that Sn is a compact subset of ℝn+1 so the earlier theorem applies to Sn. Consider θ